Question

One side of an equilateral triangle is 24 cm. The midpoints of its sides are joined to form another triangle whose midpoints are in turn joined to form still another triangle. This process continues indefinitely. Find the sum of the perimeters of all the triangles.

• A. 144 cm
• B. 72 cm
• C. 536 cm
• D. 676 cm

Answer 1

The Correct Option is A

An equilateral triangle with side length \(24 \, \text{cm}\) is continuously subdivided by joining midpoints of its sides. We aim to find the sum of perimeters of all such triangles formed. Initially, in an equilateral triangle with side \(a\), the perimeter is \(3a\). For the first triangle, \(a_1 = 24 \, \text{cm}\), so \(P_1 = 3 \times 24 = 72 \, \text{cm}\).

Joining midpoints of the sides of the previous equilateral triangle forms a smaller equilateral triangle, with its side length being half of the original. Hence, \(a_2 = \frac{a_1}{2} = \frac{24}{2} = 12 \, \text{cm}\) and \(P_2 = 3 \times 12 = 36 \, \text{cm}\).

This process continues indefinitely. Each subsequent triangle's side is half of the prior, conforming to: \(a_n = \frac{a_{n-1}}{2}\), thus perimeter \(P_n = 3 \times a_n\).

The perimeter sum is expressed as a geometric series: \[S = P_1 + P_2 + P_3 + \cdots = 72 + 36 + 18 + \cdots\]

Here, \(S\) is the sum of an infinite geometric series with first term \(a = 72\) and common ratio \(r = \frac{1}{2}\). The formula for this sum is: \[S = \frac{a}{1 - r} = \frac{72}{1 - \frac{1}{2}} = \frac{72}{\frac{1}{2}} = 72 \times 2 = 144 \, \text{cm}\]

Thus, the sum of the perimeters of all triangles is \(144 \, \text{cm}\).